Question

$2.10$ g mixture of $NaHC{O}_3$ and $KCl{O}_3$ requires $100$ mL of $0.1N$ $HCl$ for complete reaction. If $x$ grams amount of residue that would be obtained on strongly heating $2.20$ g of the same mixture then, determine the value of $1000x$.

• A. $1358$
• B. $1300$
• C. $1380$
• D. None of the above

Answer 1

The correct option is A $1358$

Only $NaHC{O}_3$ reacts with $0.1N$ $HCl$.

Let $x$ g be the weight of $NaHC{O}_3$ and $(2.1-x)$ g be the weight of $KCl{O}_3$.
Milliequivalents of $NaHC{O}_3=100\times 0.1=10$
Weight of $NaHC{O}_3=10\times {10}^{-3}\times 84=0.84g$
Weight of $KCl{O}_3=2.1-0.84=1.26g$
$2NaHC{O}_3\stackrel{\Delta }{\to }N{a}_{2}C{O}_3+H_{2}O\uparrow +C{O}_2\uparrow$
$2$ mol of $NaHC{O}_3=2\times 82$ g
1 mol of $N{a}_{2}C{O}_3=106$ g
$2KCl{O}_3\stackrel{\Delta }{\to }2KCl+3O_2\uparrow$
The residue will have $N{a}_{2}C{O}_3$ and $KCl$.
Weight of residue from $2.1$ g of mixture$=\frac{53}{84}\times 0.84+\frac{74.5}{122.5}\times 1.26$$=0.53+0.766=1.296$ g
Weight of residue from $2.2$ g of the mixture $=\frac{1.296}{2.1}\times 2.2=1.358$ g

Hence, the value of $1000x$ is $1358$.